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Numerade Educator



Problem 1 Medium Difficulty

Given that
$$ \displaystyle \lim_{x\to a} f(x) = 0 $$ $$ \displaystyle \lim_{x\to a} g(x) = 0 $$ $$ \displaystyle \lim_{x\to a} h(x) = 1 $$
$$ \displaystyle \lim_{x\to a} p(x) = \infty $$ $$ \displaystyle \lim_{x\to a} q(x) = \infty $$
which of the following limits are indeterminate forms? For those that are not an indeterminate form, evaluate the limit where possible

(a) $ \displaystyle \lim_{x\to a} \frac{f(x)}{g(x)} $
(b) $ \displaystyle \lim_{x\to a} \frac{f(x)}{p(x)} $
(c) $ \displaystyle \lim_{x\to a} \frac{h(x)}{p(x)} $
(d) $ \displaystyle \lim_{x\to a} \frac{p(x)}{f(x)} $
(e) $ \displaystyle \lim_{x\to a} \frac{p(x)}{q(x)} $


(a) $\lim _{x \rightarrow a} \frac{f(x)}{g(x)}$ is an indeterminate form of type $\frac{0}{0}$
(b) $\lim _{x \rightarrow a} \frac{f(x)}{p(x)}=0$ because the numerator approaches 0 while the denominator becomes large.
(c) $\lim _{x \rightarrow a} \frac{h(x)}{p(x)}=0$ because the numerator approaches a finite number while the denominator becomes large.
(d) If $\lim _{x \rightarrow a} p(x)=\infty$ and $f(x) \rightarrow 0$ through positive values, then $\lim _{x \rightarrow a} \frac{p(x)}{f(x)}=\infty .\left[\text { For example, take } a=0, p(x)=1 / x^{2}\right.$
and $f(x)=x^{2} .$ ] If $f(x) \rightarrow 0$ through negative values, then $\lim _{x \rightarrow a} \frac{p(x)}{f(x)}=-\infty .$ [For example, take $a=0, p(x)=1 / x^{2}$
and $f(x)=-x^{2}$.] If $f(x) \rightarrow 0$ through both positive and negative values, then the limit might not exist. [For example,
take $\left.a=0, p(x)=1 / x^{2}, \text { and } f(x)=x .\right]$
(e) $\lim _{x \rightarrow a} \frac{p(x)}{q(x)}$ is an indeterminate form of type $\frac{\infty}{\infty}$.


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Video Transcript

So part of using look tell Israel is understanding uh the indeterminate forms. So, first we have um the limit of F of X over G of X as X approaches A and that would be 0/0. So we know that 0/0 is an indeterminate form. So that would not be allowed. Then we want F of X over P of X. So it's going to be zero over infinity. So if we have zero over infinity, we know that that was just equal zero. So the limit would equal zero. That would not be an indeterminate form. Um then for part C we would have HIV X over P of X. That be one over infinity. We know that one over infinity is zero because it's an infinitely large number um in the denominator. So I would just go to zero. Then we want to look at P of X over ffx that's going to be infinity over zero. We know that that is not allowed. That would be an indeterminant form, and then P of X over Q of X would be infinity over infinity. Mm hmm. All right. Let me Which is also known as an indeterminate forms. That's why it says undefined. Um, and those would be all the different answers.